interest paid over a three year period. Solution: 8 = 0 . 08, and n - - PDF document

Chapter 7: Simple and Compound Interest

1

7.1 Introduction ةـمدـقم

 Interest rate calculations arise in a variety of business applications, and affect all of us in our personal and professional lives.  People earn interest on sums they have invested in savings accounts.  Many home owners pay interest on money they have borrowed for mortgages, personal loans.

7.2 Simple Interest ةطيسبلا ةدـئافلا

 An amount of money which is invested, or borrowed, is called the principal.  The amount of interest depends upon the principal, the period of time over which the interest can accumulate, and the interest rate.  The interest rate is usually expressed as a percentage per period of time, for example 6% per annum.

SET 2

Chapter 7

Simple and Compound Interest

لاةـبكرملا ةدـئافلا و ةـطيسبلا ةدـئاف

2

Chapter 7: Simple and Compound Interest

The simplest form interest can take is called simple interest and is given by the formula:

I = P.i .n (Simple interest formula)

Where: I = the interest earned or paid P = the principal i = the interest rate per time period, and n = the number of time periods over which the interest accumulates Example 1. If an amount of £4000 is invested in a savings account at an interest rate of 8% per year, calculate the simple interest paid over a three year period.

Solution:

P = £4000, i = 8% =

8 100 = 0.08, and n = 3. So,

I = Pin I = (4000)(0.08)(3) I = £960 This is equivalent to earning interest of £960 ÷ 3 = £320 in each of the three years.

7.3 Compound Interest ةـبكرملا ةدـئافلا

 In the previous example we can imagine earning interest of £320 each year, and withdrawing it immediately it is paid.  However, in practice the interest earned at the end of each year can be left in the savings account so that it too earns interest.  This leads to the concept of compound interest.  In compound interest calculations interest earned, or due, for each period, is added to the principal.  If a principal P is invested at a rate i per time period, it accrues to an amount S after n time periods given by:

S = P (1 + i)n (Compound interest formula)

Chapter 7: Simple and Compound Interest

3 Example 2. Calculate the amount to which £4000 grows at

an interest rate of 8% per annum, for three years, if all the interest earned is reinvested. Solution:

The interest rate is 8% =

8 100

• r 0.08

S = P (1+ i)

n

S = 4000(1 + 0.08)

3

S = £5038.85

Example 3. A person invests $1500 in a two-year bond paying 4.5%

interest per year. Money is left in the account for the whole of the two-year period. Assuming compound interest, what amount will be in the account at the end

• f the two-year period?

Solution:

Amount invested = P = $1,500 Interest rate = i = 0.045 p.a. (4.5%) Number of time periods = n = 2 S = P(1+ i)n S = 1500(1+0.045)

2

S = 1500(1.045)

2

S = $1638.04 Amount paid out after two years is $1,638.04

Example 4. Assuming compound interest, calculate the interest earned

by investing: (a) OMR 500 for 3 years at 4% per annum (b) OMR 400 for 2 years at 0.5% per month (c) OMR 300 for 1 year at 2% per half-year.

Solution:

(a) P = OMR 500, i = 0.04 and n = 3 S = P(1+ i)

n

S = 500(1+ 0.04)

3

S = 500(1.04)

3

S = OMR 562.43 Interest earned = 562.43 – 500 = OMR 62.43

4

Chapter 7: Simple and Compound Interest

(b) P = OMR 400, i = 0.005 and n = 24 (2 years = 24 months) S = P(1+ i)

n

S = 400(1+ 0.005)

24

S = 400(1.005)

24

S = OMR 450.86 Interest earned = 450.86 – 400 = OMR 50.86 (c) P = OMR 300, i = 0.02 and n = 2 (1 year = 2 half-years) S = P(1+ i)

n

S = 300(1+ 0.02)

2

S = 300(1.02)

2

S = OMR 312.12 Interest earned = 312.12 – 300 = OMR 12.12

Example 5. Rashid has OMR 10,000 to invest for one year and is

considering three different accounts: (a) A one year bond offering 4% per annum (b) An account offering 0.35% per month (c) An account offering 2.1% per half-year. He does not need the interest until the end of the year. Assuming compound interest, into which account should he invest his money to maximize the interest?

Solution:

Using S = P(1+ i)

n in each case gives:

(a) P = OMR 10,000, i = 0.04 and n = 1 Then, S = 10000(1+ 0.04)

1

S = 10000(1.04) S = OMR 10400 Interest earned = 10400 – 10000 = OMR 400 (b) P = OMR 10,000, i = 0.0035 and n = 12 S = 1000(1+ 0.0035)

12

S = 10000(1.0035)

12

S = OMR 10428.2 Interest earned = 10428.2 – 10000 = OMR 428.2  (c) P = OMR 10,000, i = 0.021 and n = 2 S = 10000(1+ 0.021)

2

Chapter 7: Simple and Compound Interest

5 S = 10000(1.021)

2

S = OMR 10424.4 Interest earned = 10424.4 – 10000 = OMR 424.4 Therefore, the account that offers 0.35% per month is the best for Rashid to invest his money.

Example 6. How much does Ali need to invest today at 2.9% compounded

semi-annually to save OMR 14,000 after 3 years?

Solution:

Amount invested today = P = ? Interest rate = i = 0.029 per half-year (2.9%) Number of time periods = n = 6 Amount in the account after 3 years = S = OMR 14,000 14000 = P (1+ 0.029)

6

14000 = P (1.029)

6

14000 = P (1.18711) P =

14000 1.18711

P = OMR 11,793.3

7.4 Continuous Compounding ةرـمتسملا ةـبكرملا ةدـئافلا

 Interest earned on an investment, or due on a loan, is usually compounded.  Compound interest was previously described when the compounding period was annual, semi- annual, monthly...etc.  On occasions, interest is compounded continuously which has the effect of increasing the amount of interest.  When interest is compounded continuously, the accrued amount at any time t is S(t) and is given by:

S(t) = P0 eit (Continuous compounding interest formula)

Where P0 = the principal invested right at the start e = the exponential constant

i = interest rate

 When using this formula, the units of time must be consistent throughout. So for example, if i is the annual interest rate, t must be measured in years.

6

Chapter 7: Simple and Compound Interest

Example 7. A principal of £1000 is invested at a constant annual rate of 8%.

Interest earned is compounded continuously. Find the accrued amount after 25 years.

Solution:

P0 = 1000, i = 0.08 and t = 25 we have:     

 2 ) 25 08 . (

1000 1000 ) 25 ( e e S £7,389.06

Example 8. Suppose that $2000 is invested at interest rate i, compounded

continuously, and grows to $2504.65 in 5 years. (a) What is the interest rate?

(b) Find the exponential growth (continuous compounding)

function S(t). (c) What will the balance be after 10 years?

(d) After how long will the $2000 be doubled?

Solution:

We know that S(5) = $2504.65. We substitute and solve for i: 045 . 5 2000 65 . 2504 ln ln 2000 65 . 2504 ln 2000 65 . 2504 2000 65 . 2504 2000 65 . 2504

5 5 5 ) 5 (

      i i e e e e

i i i i

The interest rate is 0.045 or 4.5%. (b) By substituting 0.045 for i in the function

it

e t S 2000 ) (  :

t

e t S

045 .

2000 ) (  (c) The balance after 10 years is:

62 . 3136 $ 2000

45 . ) 10 ( 045 .

2000 ) 10 (

 

 e e S (a) At t = 0, S(0) = P0 = $2000. So, the exponential growth function is of the form:

it

e t S 2000 ) ( 

Chapter 7: Simple and Compound Interest

7 (d) To find the doubling time T: S(T) = 2  P0 = 2  $2000 = $4000 Solve for T:

045 . 2 ln 045 . 2 ln ln 2 ln 2

045 . 045 . 045 . 045 .

2000 4000 2000 4000

   

 

T T

T T T T

e e e e T = 15.4 years Thus, the original investment of $2000 will double in about 15.4 years.